Efficiently Computing SSA

1
Efficiently Computing SSA
2
Static Single Assignment form

• Each name is defined exactly once
• Each use refers to exactly one name
• Why use SSA form?

3
Joins

• What to do when two different values meet on the
  Control Flow Graph?
• F-Functions
• A F-function is a special kind of copy that
  selects one of its parameters

4
Sparse Representation

• Only want to propagate facts where relevant
• Dont care about the rest
• Use the SSA Graph

5
SSA Graph

• Add edges from definitions to uses

6
Example

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

7
Example
Number existing defns

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

i1 j k l 1 repeat if (p) then begin
j i if Q then l 2 else l 3
k k 1 end else k k 2 print
(i,j,k,l) repeat if R then l l 4
until S i2 i 6 until T
8
Example
Add ?-functions where needed
i1 j k l 1 repeat i3 ?() if (p)
then begin j i if Q then l 2
else l 3 k k 1 end else k k
2 print (i,j,k,l) repeat if R then l
l 4 until S i2 i 6 until T

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

9
Example
Fill in the use numbers --- Then repeat for other
variables
i1 j k l 1 repeat i3 ?(i1,i2) if
(p) then begin j i3 if Q then l
2 else l 3 k k 1 end else k
k 2 print (i3,j,k,l) repeat if R
then l l 4 until S i2 i3 6 until T

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

10
Example

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

i1 j1 k1 l1 1 repeat i3 f(i1,i2) j2
f(j1,j4) k2 f(k5,k1) l2 f(l9,l1)
if (p) then begin j3 i2 if Q then
l3 2 else l4 3 l5 f(l3,l4)
k3 k2 1 end else k4 k2 2 j4
f(j3,j2) k5 f(k3,k4) l6 f(l2,l5)
print (i3,j4,k5,l6) repeat l7
f(l9,l6) if R then l8 l7 4 l9
f(l7,l8) until S i2 i3 6 until T
11
Example - CFG
Entry
1
2

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

3
7
4
5
6
8
9
10
11
12
Exit
12
Constructing SSA (Naively)

• Insert F-functions at every join for every name
• Solve Reaching Definitions
• Rename each use to the definition that reaches it

13
Constructing SSA (Naively)

• Why is the previous bad?
• Too many ?-functions!
• Can we do better?

14
Efficiently Constructing SSA

• Perform Control-flow analysis
• Insert ?-Functions
• Rename Values
• However where do we put the
• ?-Functions?

15
Formalizing ? placement

• We need a F function at node Z if
• Two non-null CFG paths that both define v
• Such that both paths start at two distinct nodes
  and end at Z

16
Dominance Frontiers Illustration
Dominated by X
Dominance Frontier of X (Not Dominated by X)
17
Dominance Frontiers

• If z is the first node we encounter on the path
  from x which x does not strictly dominate, z is
  in the dominance frontier of x
• For this to happen, there is some path from node
  x to z, x ? ? y ? z
• where (x SDOM y) but not (x SDOM z).

18
Example
1

• DF(1)
• DF(2)
• DF(3)
• DF(4)
• DF(5)
• DF(6)
• DF(7)

7
6
6
1, 7
7
Ø
19
Computing Dominance Frontiers

• Two components to DF(X)
• DFlocal(X) Y? succ(X) Xgt Y
• Any child of X not (strictly) dominated by X is
  in DF(X)
• Let Z be such that idom(Z) X
• idom(Z) is the parent of Z in the dominator tree
• DFup(Z) Y? DF(Z) XgtY
• Nodes from DF(Z) that are not strictly dominated
  by X are also in DF(X)

20
Algorithm

• Let SDOM(X) Y XgtY
• For each Z such that idom(Z) X do
• DF(X) DFlocal(X) ? (DF(Z) - SDOM(X))
• I.e., DF(X) DFlocal(X) ? DFup(Z)

21
SSA f-placement

• For each variable M in the program
• Set Worklist DM, the set of CFG nodes that
  contain assignments to variable M.
• While items left in the Worklist
• remove some node X from the Worklist
• for all W in DF(X), if W has never been in M's
  Worklist, add a f-term at W. Place W into the
  Worklist.

22
CFG for example Variable k
Entry
1
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
2
7
4
5
7
3
8
6
4
5
6
9
8
11
10
9
10
12
11
12
Exit
23
1 dominates the entire graph

• Worklist 1,6,7 location of ks assignments
• Compute DF(1) 2,3,4,5,6,7,8,9,10,11,12,Exit
  2,3,4,5,6,7,8,9,10,11,12,Exit
• Remember that DF(x) (SUCC(DOM-1(x)) SDOM-1(x)
  where SUCC(x) is set of successors of x in the
  CFG)

Entry
1
2
3
7
4
5
6
8
9
10
11
12
Exit
24

• Worklist 6,7
• Compute DF(6) 8 8
• put f in node 8
• Remember that DF(x) (SUCC(DOM-1(x)) SDOM-1(x)
  where SUCC(x) is set of successors of x in the
  CFG)

Entry
1
2
3
7
4
5
node 6 only dominates itself
6
f
8
9
10
11
12
Exit
25

• Worklist 7,8
• Compute DF(7) 8 8
• node 8 already in worklist
• Remember that DF(x) (SUCC(DOM-1(x)) SDOM-1(x)
  where SUCC(x) is set of successors of x in the
  CFG)

Entry
1
2
3
7
4
5
node 7 only dominates itself
6
f
8
9
10
11
12
Exit
26

• Worklist 8
• Compute DF(8) 2,9,10,11,12, exit
  9,10,11,12,exit 2
• put f in node 2
• Remember that DF(x) (SUCC(DOM-1(x)) SDOM-1(x)
  where SUCC(x) is set of successors of x in the
  CFG)

Entry
1
f
2
3
7
4
5
node 8 dominates 8,9,10,11,12,exit
6
f
8
9
10
11
12
Exit
27

• Worklist 2
• Compute DF(2) 2-12, exit 3-12,exit 2
• Remember that DF(x) (SUCC(DOM-1(x)) SDOM-1(x)
  where SUCC(x) is set of successors of x in the
  CFG)

Entry
1
f
2
3
7
4
5
node 2 dominates 2-12,exit
6
f
8
9
10
11
12
Exit
28
CFG for example Variable k
Entry
k 1
1
2
k f(k,k)
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

7
4
5
k k 2
6
k k 1
8
k f(k,k)
9
10
11
12
Exit
29
Linking it all together

• Initially, for each variable V
• C(V) 0 / counter /
• S(V) Empty / stack for each
  variable/
• Other important information
• RHS(A) set of variables used on right side of
  assignment
• LHS(A) set of variables assigned to on left
  side
• WhichPred(X,Y) j where Y is the jth predecessor
  of X in CFG
• Children(X) children in the IDOM tree
• Search(entry) will do renaming of all variables
  in program.

30

• Search (X CFG node)
• for each statement A in X
• If A is an ordinary assignment statement
• For each v in RHS(A) replace the use of v with
  vi, where i top(S(v))
• For each v in LHS(A)
• i C(v) replace v by vi
• push i on S(v) C(v)
• end
• end
• For each Y in Succ(X) do
• J WhichPred(X,Y) / predecessors have unique
  wrt successor /
• For each f-function F in Y
• Replace j-th operand v in RHS(F) by vi where i
  Top(S(v))
• End
• end
• For each Y in Children() call Search() end
• For each assignment A in X
• For each v in LHS(A) pop S(v) end
• end

top(S(v)) will hold the current for variable v
New assignment update the counter and push
31

• Search (X CFG node)
• for each statement A in X
• If A is an ordinary assignment statement
• For each v in RHS(A) replace the use of v with
  vi, where i top(S(v))
• For each v in LHS(A)
• i C(v) replace v by vi
• push i on S(v) C(v)
• end
• end
• For each Y in Succ(X) do
• j WhichPred(X,Y) / predecessors have unique
  wrt successor /
• For each f-function F in Y
• Replace j-th operand v in RHS(F) by vi where i
  Top(S(v))
• End
• end
• For each Y in Children() call Search() end
• For each assignment A in X
• For each v in LHS(A) pop S(v) end
• end

Look at successors of this node in CFG
Resolve any ? functions in successors
32

• Search (X CFG node)
• for each statement A in X
• If A is an ordinary assignment statement
• For each v in RHS(A) replace the use of v with
  vi, where i top(S(v))
• For each v in LHS(A)
• i C(v) replace v by vi
• push i on S(v) C(v)
• end
• end
• For each Y in Succ(X) do
• J WhichPred(X,Y) / predecessors have unique
  wrt successor /
• For each f-function F in Y
• Replace j-th operand v in RHS(F) by vi where i
  Top(S(v))
• End
• end
• For each D in Children(X) call Search(D) end
• For each assignment A in X
• For each v in LHS(A) pop S(v) end
• end

Look at all nodes dominated by X
33

• Search (X CFG node)
• for each statement A in X
• If A is an ordinary assignment statement
• For each v in RHS(A) replace the use of v with
  vi, where i top(S(v))
• For each v in LHS(A)
• i C(v) replace v by vi
• push i on S(v) C(v)
• end
• end
• For each Y in Succ(X) do
• J WhichPred(X,Y) / predecessors have unique
  wrt successor /
• For each f-function F in Y
• Replace j-th operand v in RHS(F) by vi where i
  Top(S(v))
• End
• end
• For each D in Children(X) call Search(D) end
• For each assignment A in X
• For each v in LHS(A) pop S(v) end
• end

Pop the var pushed earlier
34
1
1
2
2
3
3
7
8
7
4
5
4 5 6
9
10 11
6
12
8
9
10
11
12
35
Just deal with k in this example
Search(1) C(k) 1 S(k) 0 Children(1) 2
k0 1
1
2
k f(k0,k)
3
7
4
5
1
k k 2
6
2
k k 1
3
7
8
8
k f(k,k)
9
4 5 6
9
10
10 11
12
11
12
Exit
36
Search(2) C(k) 2 S(k) 0,1 None of the
successors have f functions Children(2)
3,7,8
k0 1
1
2
k1 f(k0,k)
3
7
4
5
1
k k 2
6
2
k k 1
3
7
8
8
k f(k,k)
9
4 5 6
9
10
10 11
12
11
12
Exit
37
Search(3) C(k) 2 S(k) 0,1 Children(3)
4,5,6 Search(4) Children(4)
Search(5) Children(5) Search(6) C(k)
3 S(k) 0,1,2 Children(6) Pop(S(k))
0,1
k0 1
1
2
k1 f(k0,k)
3
7
4
5
k k 2
6
1
k2 k1 1
8
k f(k2,k)
2
9
3
7
8
10
4 5 6
9
11
10 11
12
Exit
12
38
Search(7) C(k) 4 S(k) 0,1,3 Children(7)
Pop(S(k)) 0,1
k0 1
1
2
k1 f(k0,k)
3
7
4
5
1
k3 k1 2
6
2
k2 k1 1
8
3
7
8
k f(k2,k3)
9
4 5 6
9
10
10 11
12
11
12
Exit
39
Search(8) C(k) 5 S(k) 0,1,4 Children(8)
9 Search(9) Children(9) 10,11 Search(10) Sea
rch(11)
k0 1
1
2
k1 f(k0,k)
3
7
4
5
k3 k1 2
1
6
k2 k1 1
2
8
k4 f(k2,k3)
3
7
8
9
4 5 6
9
10
10 11
11
12
Exit
12
40
Search(12) C(k) 5 S(k) 0,1,4
k0 1
1
2
k1 f(k0,k4)
3
7
4
5
k3 k1 2
1
6
2
k2 k1 1
8
k4 f(k2,k3)
3
7
8
9
4 5 6
9
10
10 11
11
12
12
Exit
41

• i j k1 l 1
• Repeat
• k2 f(k1,k5)
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k3 k2 1
• end else k4 k2 2
• k5 f(k3,k4)
• print (i,j,k5,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

• i j k l 1
• Repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

42
Consider L
Entry
l
1
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

7
4
5
l
l
6
print l
8
9
10
l l
11
12
Exit
43
Computing the DF
Entry
l
1
l f
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l
l
7
3
8
6
l f
l f
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
44

• i j k l 1
• repeat
• l f(l,l)
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• l f(l,l)
• k k 1
• end else k k 2
• l f(l,l)
• print (i,j,k,l)
• repeat
• l f(l,l)
• if R then l l 4
• l f(l,l)
• until S
• i i 6
• until T

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

45
Adding the numberssearch(1)
Entry
l1
1
l f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l
l
7
3
8
6
l f
l f
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
46
Adding the numberssearch(2)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l
l
7
3
8
6
l f
l f
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
47
Adding the numberssearch(3)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l
l
7
3
8
6
l f
l f
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
48
Adding the numberssearch(4)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l
l3
7
3
8
6
l f(l3)
l f
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
49
Adding the numberssearch(5)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l f(l3,l4)
l f
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
50
Adding the numberssearch(6)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l5 f(l3,l4)
l f(l5
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
51
Adding the numberssearch(7)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l5 f(l3,l4)
l f(l5,l2)
4
5
6
9
print l
8
9
11
10
l f
10
12
l l
l f
11
12
Exit
52
Adding the numberssearch(8)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l5 f(l3,l4)
l6 f(l5,l2)
4
5
6
9
print l6
8
9
11
10
l f(l6)
10
12
l l
l f
11
12
Exit
53
Adding the numberssearch(9)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l5 f(l3,l4)
l6 f(l5,l2)
4
5
6
9
print l6
8
9
11
10
l7 f(l6)
10
12
l l
l f(l7)
11
12
Exit
54
Adding the numberssearch(10)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l5 f(l3,l4)
l6 f(l5,l2)
4
5
6
9
print l6
8
9
11
10
l7 f(l6)
10
12
l8 l7
l f(l7,l8)
11
12
Exit
55
Adding the numberssearch(11)
Entry
l1
1
l2 f(l1)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l5 f(l3,l4)
l6 f(l5,l2)
4
5
6
9
print l6
8
9
11
10
l7 f(l6,l9)
10
12
l8 l7
l9 f(l7,l8)
11
12
Exit
56
Adding the numberssearch(12)
Entry
l1
1
l2 f(l1,l9)
2
3

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

1
7
4
5
2
l4
l3
7
3
8
6
l5 f(l3,l4)
l6 f(l5,l2)
4
5
6
9
print l6
8
9
11
10
l7 f(l6,l9)
10
12
l8 l7
l9 f(l7,l8)
11
12
Exit
57

• i j k l1 1
• repeat
• l2 f(l1,l9)
• if (p) then begin
• j i
• if Q then l3 2
• else l4 3
• l5 f(l3,l4)
• k k 1
• end else k k 2
• l6 f(l5,l2)
• print (i,j,k,l6)
• repeat
• l7 f(l6,l9)
• if R then l8 l7 4
• l9 f(l7,l8)
• until S
• i i 6
• until T

• i j k l 1
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k k 1
• end else k k 2
• print (i,j,k,l)
• repeat
• if R then l l 4
• until S
• i i 6
• until T

58
SSA Deconstruction

• At some point, we need executable code
• Few machines implement ? operations
• Need to fix up the flow of values
• Basic idea
• Insert copies ?-function preds
• Simple algorithm
• Works in most cases
• Adds lots of copies
• Most of them coalesce away

59

• i j k0 l 1
• k1 k0
• repeat
• if (p) then begin
• j i
• if Q then l 2
• else l 3
• k2 k2 1
• k4 k2
• end else
• k3 k1 2
• k4 k3
• print (i,j,k4,l)
• repeat
• if R then l l 4
• until S
• i i 6
• k1 k4
• until T

k0 1 k1 k0
1
2
k1 f(k0,k4)
3
7
4
5
k3 k1 2 k4 k3
6
k2 k1 1 k4 k2
8
k4 f(k2,k3)
9
10
11
12
Exit
k1 k4